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CIE 9709 2023 June Paper 62 Q5

A Level / CIE / S2

CIE 9709 2023 June Paper 62 Paper · Question 5

题目

Problem

(a) Two random variables XX and YY have the independent distributions N(7,3)\mathrm{N}(7,3) and N(6,2)\mathrm{N}(6,2) respectively. A random value of each variable is taken.

Find the probability that the two values differ by more than 22.

(5)

(b) Each candidate’s overall score in a science test is calculated as follows. The mark for theory is denoted by TT, the mark for practical is denoted by PP, and the overall score is given by T+1.5PT+1.5P. The variables TT and PP are assumed to be independent with distributions N(62,158)\mathrm{N}(62,158) and N(42,108)\mathrm{N}(42,108) respectively. You should assume that no continuity corrections are needed when using these distributions.

(i) A pass is awarded to candidates whose overall score is at least 9090.

Find the proportion of candidates who pass.

(5)

(ii) Comment on the assumption that the variables TT and PP are independent.

(1)
题目中文翻译

(a) 两个随机变量 XX 和 YY 相互独立，分别服从 N(7,3)\mathrm{N}(7,3) 和 N(6,2)\mathrm{N}(6,2) 分布。分别随机选取这两个变量的一个取值。

求这两个值相差超过 22 的概率。

(5)

(b) 在一次科学考试中，每名考生的总分按下述方法计算。理论部分的分数记为 TT，实操部分的分数记为 PP，总分为 T+1.5PT+1.5P。假设 TT 和 PP 相互独立，分别服从 N(62,158)\mathrm{N}(62,158) 和 N(42,108)\mathrm{N}(42,108) 分布。假设使用这些分布时无需进行连续性修正。

(i) 总分至少为 9090 的考生通过考试。

求通过考试的考生所占的比例。

(5)

(ii) 评价 TT 和 PP 相互独立这一假设。

(1)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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相差超过2表示差大于2或小于负2，先建立正态差分布，再加两个不相交尾部。

答题过程

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By independence,

D=X−Y.D=X-Y.

Then

D∼ N(7−6, 3+2)= N(1,5).\begin{align*} D\sim&\,\mathrm{N}(7-6,\ 3+2)\\[4mm] =&\,\mathrm{N}(1,5). \end{align*}

The required event is ∣D∣>2|D|>2, so

P(∣D∣>2)= P(D>2)+P(D<−2)= 1−Φ(2−15) +Φ(−2−15)= 0.417217…= 0.417(3 s.f.).\begin{align*} \mathrm{P}(|D|>2)=&\,\mathrm{P}(D>2)+\mathrm{P}(D<-2)\\[4mm] =&\,1-\Phi\bigg(\frac{2-1}{\sqrt5}\bigg)\\[4mm] &\,+\Phi\bigg(\frac{-2-1}{\sqrt5}\bigg)\\[4mm] =&\,0.417217\ldots\\[4mm] =&\,\boxed{0.417}\quad\text{(3 s.f.).} \end{align*}

(b)(i)

解法一

思路

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总分是T加1.5P，独立性使方差按系数平方相加。按题目要求不使用连续性修正。

答题过程

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Let S=T+1.5PS=T+1.5P be the overall score. Then

E(S)=62+1.5(42)=125,\mathrm{E}(S)=62+1.5(42)=125,

and, by independence,

Var(S)=158+(1.5)2(108)=401.\mathrm{Var}(S)=158+(1.5)^2(108)=401.

Thus S∼N(125,401)S\sim\mathrm{N}(125,401). Without a continuity correction, as instructed,

P(S≥90)= 1−Φ(90−125401)= 0.959752…= 0.960(3 s.f.).\begin{align*} \mathrm{P}(S\geq90)=&\,1-\Phi\bigg(\frac{90-125}{\sqrt{401}}\bigg)\\[4mm] =&\,0.959752\ldots\\[4mm] =&\,\boxed{0.960}\quad\text{(3 s.f.).} \end{align*}

(b)(ii)

解法一

思路

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理论与实验成绩可能共同反映科学能力，理论表现较好者往往实验也较好，因此独立假设不太合理。

答题过程

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The independence assumption is unlikely to be realistic. A candidate who performs well in theory is also likely to perform well in practical work.